$g(x) = x^{3}-7x^{2}+2(f(x))$ $f(x) = 6x+4(h(x))$ $h(t) = t-7$ $ h(f(6)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(6)$ . Then we'll know what to plug into the outer function. $f(6) = (6)(6)+4(h(6))$ To solve for the value of $f$ , we need to solve for the value of $h(6)$ $h(6) = 6-7$ $h(6) = -1$ That means $f(6) = (6)(6)+(4)(-1)$ $f(6) = 32$ Now we know that $f(6) = 32$ . Let's solve for $h(f(6))$ , which is $h(32)$ $h(32) = 32-7$ $h(32) = 25$